Using Eisenstein integers $\mathbb{Z}[w]$, I found that its units are $\pm1,\pm w,\pm(1+w)$, etc. Where is the mistake?

Question

I'm totally confused with that: I studied the groups of order 6 and according for example to this answer they are isomorphic to $\mathbb{Z}_6$ so they are cyclic or isomorphic to $S_3$ so they are non-abelian. The problem is that by working with the Eisenstein integers $\mathbb{Z}[w]$ which is a sub-ring of $(\mathbb C,+,\times)$ I found that its units are $\pm1,\pm w,\pm(1+w)$ and we know that this set of units is a sub-group of $(\mathbb C^*,\times)$ of order 6 but it's not cyclic because neither element has the order 6 so it's not isomorphic to $\mathbb Z_6$ but also it's abelian so not isomorphic to $S_3$. Can you explain me where is the mistake?

Answer 1

The group is cyclic. The element $-\omega$ has order six.

More generally, if $F$ is a field then there is a theorem that states that any finite subgroup of $F^*$ is cyclic.

Answer 2

The units in $\mathbb Z[\omega]$ are $U=\{\pm 1,\pm\omega, \pm(1+\omega)\}$ where $w^3=1$, and thus $U$ is a subgroup of $\mathbb C^\times$ with $6$ elements, and so is an abelian group of order $6$.

Now if $G = \mathbb Z/6\mathbb Z$ then it has

• 1 element of order 1: $0+6\mathbb Z$;
• 1 element of order $2$: $3+6\mathbb Z$;
• 2 elements of order $3$: $\pm 2+6\mathbb Z$;
• 2 elements of order $6$: $\pm 1+6\mathbb Z$.

But clearly $1$ has order $1$ while $-1$ has order $2$ and $\omega$ has order $3$. Since $1+\omega+\omega^2=0$ we have $\omega(-1-\omega)=1$, and so $-1-\omega = \omega^{-1}$ also has order $3$, and so if $U$ is cyclic of order $6$ it must be that $-\omega$ and $1+\omega$ have order $6$. This can be checked by a direct calculation, but since $\omega$ has order $3$ and $-1$ clearly has order $2$, the fact that $-\omega$ has order $6$ follows from the following easy general claim:

Claim: If $x_1$ and $x_2$ are elements of any abelian group $\Gamma$ with orders $d_1$ and $d_2$ respectively and $d_1,d_2$ are coprime, then $x_1x_2$ has order $d_1d_2$.

Proof: Let $C_1 = \langle x_1 \rangle$ and $C_2 = \langle x_2 \rangle$ be the cyclic subgroups of $\Gamma$ generated by $x_1$ and $x_2$ respectively. Then if $y \in C_1\cap C_2$ and $e= \text{ord}(y)$ is the order of $y$, by Lagrange's theorem we have $e\mid d_1$ and $e \mid d_2$, so that when $d_1,d_2$ are coprime, $C_1\cap C_2 = \{e\}$.

Now if $k\in \mathbb Z$, since $\Gamma$ is abelian, we have $(x_1x_2)^k = x_1^kx_2^k$ hence $(x_1x_2)^k=e$ if and only if $x_1^k=x_2^{-k}$. But then as $C_1 \cap C_2 =\{e\}$ it follows that $x_1^k=e=x_2^{k}$. But then $d_1\mid k$ and $d_2 \mid k$, so that $\text{l.c.m}(d_1,d_2)\mid k$ and since $d_1,d_2$ are coprime, $\text{l.c.m.}(d_1,d_2)= d_1d_2$ and it follows that the order of $x_1.x_2$ is $d_1d_2$ as claimed. $\blacksquare$

Finally, on Mark’s comment above that a finite subgroup of the multiplicative group of any field is always cyclic, the special case where $F=\mathbb C$ is not hard to prove: if $H \leq \mathbb C^\times$ is a finite subgroup of the multiplicative group $\mathbb C^\times$, and $|H|=n$, then by Lagrange's theorem, $h^n=1$ for all $h \in H$. But then every element of $H$ is a solution to the equation $t^n-1=0$. It follows that $H \subseteq\{z \in \mathbb C: z^n-1=0\}=: \mu_n$. But if $\omega_n := \exp(2\pi i/n)$ then $\mu_n = \{\omega^k: 0\leq k \leq n-1\}$ and hence it is a cyclic group of order $n$, and as $H \subseteq \mu_n$ and $|H|=\mu_n$ we must have $H=\mu_n$ and hence $H$ is cyclic as required.

[If $F$ is an arbitrary field the set $\mu_n(F) = \{z \in F: z^n-1=0\}$ is always an abelian group which contains any subgroup of $F^\times$ of order $n$, and as a polynomial has at most as many roots as its degree, it follows that if $H$ is a subgroup of $F^\times$ of order $n$, then $H=\mu_n(F)$ and $|\mu_n(F)|=n$, but it is not immediately clear that $\mu_n(F)$ must be cyclic.]